Let's solve the problem step by step using **Ampère's Law**: ## Step 1: Applying Ampère's Law Ampère's Law states: \[ \oint \vec{B} \cdot d\vec{l} = \mu_ I_{\text{enc}} \] where \( I_{\text{enc}} \) is the net current enclosed by the path (taking into account the direction: out of the page is positive, into the page is negative). --- ### Path 1: - Path 1 encloses four conductors. - Let’s count the direction of the currents: - If out of the page (\(\odot\)): **positive** - If into the page (\(\otimes\)): **negative** - Suppose Path 1 encloses 2 out and 2 in (as shown). - Net current \( I_{\mathrm{enc,1}} = 9.69 + 9.69 - 9.69 - 9.69 = \) A **Explanation:** The net current enclosed is zero since the contributions cancel out. --- ### Path 2: - Path 2 also encloses four conductors. - Suppose Path 2 encloses 3 out and 1 in (as shown). - Net current \( I_{\mathrm{enc,2}} = 9.69 + 9.69 + 9.69 - 9.69 = 9.69 \times 2 = 19.38 \) A **Explanation:** The net current is the sum of the three outwards and one inward, totaling \( 19.38 \) A. --- ## Step 2: Calculating the Line Integral Now, use Ampère's Law for each path: ### (a) Path 1 \[ \oint \vec{B} \cdot d\vec{l}_1 = \mu_ \times = \] **Explanation:** No net current is enclosed; thus, the line integral is **zero**. ### (b) Path 2 \[ \oint \vec{B} \cdot d\vec{l}_2 = \mu_ \times 19.38 \] where \(\mu_ = 4\pi \times 10^{-7}\) T·m/A. \[ \oint \vec{B} \cdot d\vec{l}_2 = 4\pi \times 10^{-7} \times 19.38 = 2.435 \times 10^{-5} \text{ T·m} \] --- ## Final Answers - **(a) Path 1:** \( \boxed{} \) T·m - **(b) Path 2:** \( \boxed{2.435 \times 10^{-5}} \) T·m